As shown in the figure, circle I is the inscribed circle of the triangle ABC, which is tangent to points D, e and F with AB, BC and Ca respectively, and the angle def = 50 degrees. Find the angle A

As shown in the figure, circle I is the inscribed circle of the triangle ABC, which is tangent to points D, e and F with AB, BC and Ca respectively, and the angle def = 50 degrees. Find the angle A

Because circle 1 is the inscribed circle of triangle ABC and is tangent to points D, e and F with AB, BC and Ca respectively
Because the angle def = 1 / 2, inferior arc DF = 50 degrees
So the inferior arc DF = 100 degrees
So arc def = 350-100 = 260 degrees
Because the angle a = 1 / 2 (arc def minor arc DF) = 1 / 2 (260-100) = 80 degrees
So the angle a is 80 degrees

It is known that, as shown in the figure, in the triangle ABC, the inscribed circle I and the edges BC, Ca and ab are tangent to points D, e and f respectively. It is proved that: ∠ FDE = 90 ° - 1 / 2 ∠ a

prove:
∵ the inscribed circle I and the edges BC, Ca and ab are tangent to points D, e and f respectively
ν BF = BD [the length of two tangent lines leading to the circle from a point outside the circle is equal]
∴∠BDF=∠BFD=（180º-∠B）÷2=90º-½∠B
∵CD=CE
∴∠CDE=∠CED=（180º-∠C）÷2=90º-½∠C
∴∠FDE=180º-∠BDF-∠CDE=180º-（90º-½∠B）-（90º-½∠C）
=½∠B+½∠C=½（∠B+∠C）
=½（180º-∠A）
=90º-½∠A

As shown in the figure, the inscribed circle I is the inscribed circle of triangle ABC, ab = 9, BC = 8, CA = 10, points D and E are points on AB and AC respectively, and De is the tangent line of inscribed circle I Find the perimeter of the triangle ade

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In this website, there is question 7. The detailed explanation is at the back

Three circles with radius root 3 are circumscribed in pairs, and each side of triangle ABC is tangent to two of them. What is the circumference of triangle ABC

∵ three circles are tangent to each other, so the ᙽ ABC of the circumscribed is an equilateral triangle, as shown in the figure,

In the right triangle ABC, the angle ACB = 90 degrees, AC = 6, BC = 8. Two circles O1 and O2 with equal radius circumscribed are inscribed on the triangle ABC

In the right triangle ABC, the angle ACB = 90 degrees, AC = 6, BC = 8, so AB = 10
tan(A/2)=sinα/(1+cosA)=0.8/(1+0.6)=1/2
tan(B/2)=sinα/(1+cosB)=0.6/(1+0.8)=1/3
AB=（2+2+3）r=10
r=10/7

In the right triangle ABC, the angle ACB = 90 degrees, AC = 6, BC = 8. Two circles O1 and O2 with equal radius circumscribed are inscribed on the triangle ABC. Find the half of these two circles Can't use the half angle formula in trigonometric function

r=10/7ba.

In △ ABC, ∠ C = 90 °, BC = 4, AC = 3, two circumscribed equal circle O1 and circle O2 are tangent to AB, AC and BC respectively, and F, h, e, G are calculated It means that the two equal circles in the right triangle of 345 are tangent to the two right angles, and the bottom of E and H is f and H

BF=BE=X,AG=AH=Y
X+Y+2R=5
Y*5/3-R*4/3=Y
X*5/4-R*3/4=X
Just solve the agenda group
Y=2R,X=3R
7R=5
R=5/7

As shown in the figure, in the right triangle ABC, the angle ACB = 90 degrees, AC = 6, BC = 8, O is the point on BC, with o as the center of the circle, OC as the radius, making a circle tangent to point d with ab, Find the radius of circle O

Let the radius be r
Let ⊙ O and ab tangent to point D, then ⊙ ODA = 90 ° (OD ⊥ AB)
∵OC=OD=R
The point R is on the angular bisector of ∠ BAC
ν Ao is the angular bisector of ∠ BAC
∴∠OAC=∠OAD
∵ ACB = ∠ ODA = 90 ° Ao is the common edge
∴△AOC≌△AOD（AAS）
∴AD=AC=6
∵AC=6 BC=8 ∠ACB=90°
∴AB=√(AC²+BC²)=10
∴BD=AB－AD=4
∵OC=R BC =8
∴OB=8－R
∵OD⊥AB
∴OB²=BD²+OD²
In other words, (8-r) 2 = 4? 2 + R
R = 3
The radius of O is 3

Known: as shown in the figure, ⊙ O1 and ⊙ O2 intersect at points a and B, line AO1 intersects ⊙ O1 at point C, intersection ⊙ O2 at point D, the extension line of CB intersects ⊙ O2 at point E, connecting de. known as CD = 8, de = 6, find the length of CE

Connect ab
∵ AC is the diameter of ⊙ O1,
∴∠ABC=90°，
∴∠ABE=90°，
∵ the quadrilateral abed is the inscribed quadrilateral of circle O2,
∴∠ADE=90°，
In RT △ CDE, CD = 8, de = 6,
∴CE=
CD2+DE2＝
82+62=10．
A: the length of CE is 10